<span>The magnitude of the rock is equal to g. After the rock is released, there are no more forces acting on it, yet gravity remains. The initial inputs, on a bridge, at an angle of 30 deg below horizontal do not matter after the release.</span>
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The answer would probably be B.
Answer:
vT = v0/3
Explanation:
The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R
Since F = F0 = F'
GMm/R² = m(v0)²/R
GM = (v0)²R (1)
Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R
Since F1 = F'
GMm/27R² = m(vT)²/3R
GM = 27(vT)²R/3
GM = 9(vT)²R (2)
Equating (1) and (2),
(v0)²R = 9(vT)²R
dividing through by R, we have
9(vT)² = (v0)²
dividing through by 9, we have
(vT)² = (v0)²/9
taking square-root of both sides,
vT = v0/3
Density = (mass) / (volume), no matter how large or small the sample is.
We can't calculate the density, because you left out the number for the volume.
Also, you didn't tell us the unit for the mass of 180.
a). If the mass is 180 grams, then the density is
(180 gm) / (volume) .
b). No matter how many pieces you crush it into, and
no matter how large or small a piece is, its density is
the same. (I just wish we knew what the density really is.)
c). A piece may have 80 grams of mass. It doesn't "weigh" 80 grams.
Since the density of the whole rock is (180 gm) / (volume),
the density of any piece of it is (180 gm) / (volume).
Multiply each side by (volume): (Density) x (volume) = 180 gm
Divide each side by (density): Volume = (180 gm) / (density)
We can't calculate the volume of an 80-gm piece, because
we don't know the density. (That's because you left the volume
out of the question.)
